On which values is the floor function continous?

Question

$ f(x) = \lfloor {x} \rfloor$

My approach for now:

$\vert x_0-x \vert < \delta \Rightarrow \vert \lfloor {x_0} \rfloor - \lfloor {x} \rfloor \vert < \epsilon$

I want to try to define $\delta$ dependent on $\epsilon$ but I do not know how to continue now. I somehow want to get the $\delta$ in this term : $\vert \lfloor {x_0} \rfloor - \lfloor {x} \rfloor \vert < \epsilon$ but how?

Answer 1

Let $k\in \Bbb Z$.

we have that

$$\forall x\in (k,k+1) \;\;f(x)=\lfloor x\rfloor=k \implies$$

$$ \lim_{x\to k^+}f(x)=k=f(k)$$

and

$$\forall x\in(k-1,k)\;\; f(x)=k-1 \implies$$ $$\lim_{x\to k^-}f(x)=k-1\neq f(k)$$

thus $f$ is discontinuous on the left of integers